# How do you graph f(x)=(1-x^2)/x using holes, vertical and horizontal asymptotes, x and y intercepts?

Jul 6, 2018

no holes
vertical asymptote at $x = 0$
$x$-intercepts at $\left(- 1 , 0\right) , \left(1 , 0\right)$
$y$-intercept is undefined
no horizontal asymptote
slant/oblique asymptote at $y = - x$

#### Explanation:

Given: $f \left(x\right) = \frac{1 - {x}^{2}}{x}$

This is a rational function: $\frac{N \left(x\right)}{D \left(x\right)} = \frac{{a}_{n} {x}^{n} + \ldots}{{b}_{m} {x}^{m} + \ldots}$

Factor the numerator:
$f \left(x\right) = - \frac{{x}^{2} - 1}{x} = - \frac{\left(x - 1\right) \left(x + 1\right)}{x}$

Holes: Since there are no terms that cancel, there are no holes.

Find vertical asymptotes: when $D \left(x\right) = 0 \implies x = 0$

Find x-intercepts by setting $f \left(x\right) = 0$:

$0 = \frac{1 - {x}^{2}}{x}$

$0 \cdot x = \left(1 - {x}^{2}\right)$

$1 - {x}^{2} = 0$

${x}^{2} = 1$

$x = \pm 1$

$x$-intercepts at $\left(- 1 , 0\right) , \left(1 , 0\right)$

Find y-intercept by setting $x = 0 \implies y =$undefined

Find horizontal asymptote by comparing m & n:

If $n < m$, horizontal asymptote is $y = 0$

If $n = m$, horizontal asymptote is $y = {a}_{n} / {b}_{m}$

If $n > m$ there is no horizontal aymptote

In the given equation, $n = 2 , m = 1 \implies n > m$ which means there is no horizontal asymptote.

If $n = m + 1$ there is a slant /oblique asymptote.

You need to use long division to find the slant/oblique asymptote :

$\text{ } - x$
$x | \overline{- {x}^{2} + 0 x + 1}$
$\text{ } \underline{- {x}^{2}}$
$\text{ } 0 x + 1$

$\text{ } \textcolor{red}{- x} + \frac{1}{x}$
$x | \overline{- {x}^{2} + 0 x + 1}$
$\text{ } \underline{- {x}^{2}}$
$\text{ } 0 x + 1$ This is the remainder

slant/oblique asymptote is at $\textcolor{red}{y = - x}$